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Fair play (Posted on 2011-12-30) Difficulty: 3 of 5
Two gamblers, A and B, divided into a few bundles the money they've recently won and agreed to toss a coin for each bundle.
A 3rd gambler, nicknamed Mathwiz remarked: "It is an even money that A will win at least five of the tosses".
How many bundles of money were there?

See The Solution Submitted by Ady TZIDON    
Rating: 3.0000 (3 votes)

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pascal solution | Comment 2 of 3 |
this solution can be derived from the property that the sum of the n'th row of pascals triangle is 2^n.  Now, if there are n piles, then we need for the sum of the 5th through n'th element of the n'th row to sum to 2^(n-1) or exactly half the total.  Now, because Pascal's triangle is symetric and the n'th row has n+1 elements (0 thru n), then we need n+1 to be even and thus n to be odd.  Since we are starting the sum at 5, then we also need for that to be the start of the second half.  So in other words, we need (n-5+1)*2=n+1
2n-10+2=n+1
n=9
thus there must be n piles. 

  Posted by Daniel on 2011-12-31 08:28:16
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